If we define the operator $\mathcal{G}f(t,x)= \frac{\partial f}{\partial t}(t,x)$, what is the value of $$ \int_0^t \mathcal{G}f(s, b(s)) ds? $$ I'm sure it's some subtlety in the Fundamental Theorem of Calculus that I'm not seeing. Thanks for the help!
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Welcome to Math.SE! Can you give more details: what kind of function (?) is $b$? And what have you tried so far? – Hrodelbert May 14 '15 at 09:08
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Sorry, I don't have 50 reputation to comment, so I'll answer it here: $b(s)$ is really a Standard Brownian Motion, $B(s)$, but that shouldn't matter much here.
The most promising approach was the Differentiating Under the Integral Sign, except that that would entail calculating the derivative of an integral that I can't calculate (that is, $\frac{d}{dt} \int_0^t f(s, B(s))ds$). If it helps, the answer to the question seems to be $-f(0, B(0))$, but I'm not sure how to get it (or even if it's right).
(Feel free to add this remark as a comment and delete this post.)
radiovac99
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Sorry I down voted by accident, I absent-mindedly thought it was a comment and meant to upvote; despite what you wrote! My vote is locked in and I can't change it, perhaps you could complain about my vote and it might be rectified. – Autolatry May 14 '15 at 09:46
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1You can edit your question with the extra information, no need to post an answer. – Henrik supports the community May 14 '15 at 09:47